PHYSICS COLLEGE

Determine the angular velocity of the merry-go-round if a jumps off horizontally in the −n direction with a speed of 2 m/s , measured relative to the merry-go-round. neglect friction and the size of each child.

Answers

Answer 1

Answer:

by angular momentum conservation we will have

angular momentum of child + angular momentum of merry go round = 0

angular momentum of child = mvR

m = mass of child

R = radius of child

v = speed = 2 m/s

now let's say moment of inertia of merry go round is I

so we will have

$m*2*R + Iw = 0$

$w = -(2mR)/(I)$

so merry go round will turn in opposite direction with above speed

Related Questions

The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 1025 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 55 volts?
At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the left. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.
If a freely falling object were somehow equipped with a speedometer, its speedreading would increase each second bya) about 15 m/s.b) a rate that depends on its initial speed.c) a variable amount.d) about 5 m/s.e) about 10 m/s.
A piano tuner hears a beat every 2.20 s when listening to a 266.0 Hz tuning fork and a single piano string. What are the two possible frequencies (in Hz) of the string? (Give your answers to at least one decimal place.)
Which describes the motion of the box based on the resulting free-body diagram?1. It is moving up with a net force of 20 N.2. It is moving to the right with a net force of 10 N.3. It is in dynamic equilibrium with a net force of 0 N.4. It is in static equilibrium with a net force of 0 N.

What is effort arm
don't say the answer of gogle

Answers

Answer:

effort arm mean the use of any work by using your hand force motion or by hand power

" A sound wave moving through air consists of alternate regions of high pressure and low pressure. If the frequency of the sound is increased, what happens, if anything, to the distance between successive high-pressure regions, and why

Answers

Answer: wavelength will reduce

Explanation: The region of low pressure is know as the rarefraction region while the region of high pressure is the compression region.

The distance between 2 successive rarefraction or compression region is known as the wavelength.

Now the question is concerned about what an increase in frequency will cause to wavelength.

The speed of sound in air is a constant and it is approximately 343 m/s.

But recall that v = fλ

By assuming a fixed value for speed (v), we have that

k = fλ

Hence, f = k/ λ

This implies that at a fixed wave speed, the wavelength and frequency have an inverse relationship.

An increase in frequency will bring about a reduction in wavelength.

true or false A permanent magnet and a coil of wire carrying a current both produce magnetic fields

Answers

Answer:

True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field

John and Linda are arguing about the definition of density. John says the density of an object is proportional to itsmass. Linda says the object's mass is proportional to its density and to its volume. Which one, if either, is correct?A. They are both wrong
B. John is correct, but Linda is wrong
C. John is wrong, but Linda is correct
D. They are both correct.
E. John must be wrong, because Linda always wins these arguments.

Answers

Answer:

They are both correct.

Explanation:

The density of an object is defined as the ratio of its mass to its volume. This implies that the density of the object is both proportional to the mass and also to the volume of the object. John only mentioned mass which is correct. Linda mentioned the second variable on which density depends which is the volume of the object.

Hence considering the both statements objectively, one can say that they are both correct.

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 162 cm , but its circumference is decreasing at a constant rate of 14.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.500 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop. Find the magnitude of the emf EMF induced in the loop after exactly time 8.00s has passed since the circumference of the loop started to decrease.

Answers

Answer:

0.00124 V

Explanation:

Parameters given:

Initial circumference = 162 cm

Rate of decrease of circumference = 14 cm/s

Magnetic field, B = 0.5 T

Time, t = 8 secs

The magnitude of the EMF induced in the loop is given as:

V = (-NBA) / t

Where N = number of turns = 1

B = magnetic field

A = area of loop

t = time taken

First, we need to find the area of the loop.

To do this, we will find the radius after the loop circumference has decreased for 8 secs.

The rate of decrease of the circumference is 14 cm/s and 8 secs has passed, which means after 8 secs, it has decreased by:

14 * 8 = 112 cm

The new circumference is:

162 - 112 = 50 cm = 0.5 m

To get radius:

C = 2 * pi * r

r = C / (2 * pi)

r = 0.5 / (2 * 3.142)

r = 0.0796 m

The area is:

A = pi * r²

A = 3.142 * 0.0796²

A = 0.0199 m²

Therefore, the EMF induced is:

V = (-1 * 0.5 * 0.0199) / 8

V = -0.00124V

This is the EMF induced in the coil.

The magnitude is |-0.00124| V = 0.00124 V.

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?

Answers

Answer:

a) $v=5.6725\,m.s^(-1)$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^(-2)$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^(-1)$

putting the values in eq. (1)

$v^2=0^2+2* 3.62 * 10$

$v=8.5088\,m.s^(-1)$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20* 8.5088$

$p=170.1764\,kg.m.s^(-1)$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)* v'=p$

$(20+10)* v'=170.1764$

$v'=5.6725\,m.s^(-1)$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$(1)/(2) (M+m).v'^2=(M+m).g.h$

$(1)/(2) (20+10)* 5.6725^2=(20+10)* 9.8* h$

$h\approx 1.6420\,m$

above the normal hanging position.

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